If $A = \frac{1}{2} \begin{bmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}$,then :

Let $A = \begin{bmatrix} 0 & 1 \\ 1 & k \end{bmatrix}$, $k \in R$ and $A^3 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. If $d = 228$, then $b + c =$

If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$,then $AB = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ and $BA = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Clearly $AB \neq BA$. Thus,matrix multiplication is not commutative. Is this statement true for all matrices?

If $A-B=\begin{bmatrix} 2 & 5 \\ 9 & 0 \end{bmatrix}$ and $A+B=\begin{bmatrix} 6 & 3 \\ -1 & 0 \end{bmatrix}$,then matrix $A =$ . . . . . .

If $A$ is a $3 \times 4$ matrix and $B$ is a matrix such that $A^{\prime}B$ and $BA^{\prime}$ are both defined,then $B$ is of the type:

Let $A = \begin{bmatrix} 2 & -2 \\ 1 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 \\ -1 & 2 \end{bmatrix}$. Then the number of elements in the set $\{(n, m) : n, m \in \{1, 2, \ldots, 10\} \text{ and } nA^n + mB^m = I\}$ is